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# The perpendicular distance of a line from the origin is 5 units,

Question:

The perpendicular distance of a line from the origin is 5 units, and its slope is -1. Find the equation of the line.

Solution:

Given: perpendicular distance from orgin is 5 units, and the slope is -1

To find : the equation of the line

Formula used :

We know that the perpendicular distance from a point $\left(x_{0}, y_{0}\right)$ to the line $a x+b y+c=0$ is given by

$D=\frac{|a x+b y+c|}{\sqrt{a^{2}+b^{2}}}$

The equation of a straight line is given by $y=m x+c$ where $m$ denotes the slope of the line.

The equation of the line is $m x-y+c=0$

Here $x_{0}=0$ and $y_{0}=0, a=m, b=-y, c=c$ and $D=5$ units

$D=\frac{|m(0)-1(0)+c|}{\sqrt{m^{2}+1^{2}}}=\frac{|c|}{\sqrt{m^{2}+1}}=\frac{c}{\sqrt{m^{2}+1}}=5$

Slope of the line $=m=-1$, Substituting in the above equation we get,

$\frac{c}{\sqrt{(-1)^{2}+1^{2}}}=5$

$\frac{c}{\sqrt{1+1}}=\frac{c}{\sqrt{2}}=5$

$c=5 \sqrt{2}$

Thus the equation of the straight line is $y=-x+5 \sqrt{2}$ or $x+y-5 \sqrt{2}=0$